📚 CAIE Year 11 Engineering: Unit Test Mock Paper Walkthrough | CAIE 11年级工程：单元测试模拟卷解析
Mock unit tests are an essential part of exam preparation for CAIE Engineering. They help you identify strengths, gaps in knowledge, and build confidence. This walkthrough unpacks typical questions you might encounter in a Year 11 end-of-unit assessment, covering mechanics, electronics, materials, manufacturing, and design processes.
单元测试模拟卷是备考CAIE工程考试的重要工具。它能帮助你找出优势、知识漏洞，并建立信心。本篇解析将剖析Year 11单元评估中常见的典型题目，涉及力学、电子学、材料、制造和设计流程等核心内容。
1. Stress & Strain Calculation | 应力与应变计算
Sample question: A steel wire of diameter 1.6 mm and original length 2.5 m is stretched by a force of 280 N. The wire extends by 4.2 mm. Calculate (a) the tensile stress in the wire, (b) the tensile strain, and (c) the Young modulus of the steel.
样题： 一根直径1.6 mm、原长2.5 m的钢丝受到280 N的拉力，伸长4.2 mm。计算(a)钢丝的拉伸应力，(b)拉伸应变，(c)钢的杨氏模量。
Step 1: Convert all units to standard SI. Diameter = 1.6 mm = 1.6 × 10⁻³ m, so radius r = 0.8 × 10⁻³ m. Extension ΔL = 4.2 mm = 4.2 × 10⁻³ m.
步骤1：将所有单位转换为国际单位制。直径 = 1.6 mm = 1.6 × 10⁻³ m，因此半径 r = 0.8 × 10⁻³ m。伸长量 ΔL = 4.2 mm = 4.2 × 10⁻³ m。
Step 2: Calculate the cross-sectional area.
A = πr² = π × (0.8 × 10⁻³)² = 2.01 × 10⁻⁶ m²
步骤2：计算横截面积。
A = πr² = π × (0.8 × 10⁻³)² = 2.01 × 10⁻⁶ m²
Step 3: Tensile stress σ = F / A = 280 N / 2.01 × 10⁻⁶ m² ≈ 1.39 × 10⁸ Pa (139 MPa).
步骤3：拉伸应力 σ = F / A = 280 N / 2.01 × 10⁻⁶ m² ≈ 1.39 × 10⁸ Pa (139 MPa)。
Step 4: Tensile strain ε = ΔL / L₀ = 4.2 × 10⁻³ m / 2.5 m = 1.68 × 10⁻³.
步骤4：拉伸应变 ε = ΔL / L₀ = 4.2 × 10⁻³ m / 2.5 m = 1.68 × 10⁻³。
Step 5: Young modulus E = σ / ε = 1.39 × 10⁸ Pa / 1.68 × 10⁻³ ≈ 8.27 × 10¹⁰ Pa (82.7 GPa). This value matches typical structural steel.
步骤5：杨氏模量 E = σ / ε = 1.39 × 10⁸ Pa / 1.68 × 10⁻³ ≈ 8.27 × 10¹⁰ Pa (82.7 GPa)。该值与典型结构钢相符。
2. Resistors in Series and Parallel | 电阻的串联与并联
Sample question: A 10 Ω resistor and a 15 Ω resistor are connected in parallel, and this combination is connected in series with a 6 Ω resistor. A 12 V battery supplies the circuit. Calculate (a) the total resistance, (b) the total current, and (c) the voltage across the 10 Ω resistor.
样题： 一个10 Ω电阻与一个15 Ω电阻并联后，再与一个6 Ω电阻串联。电路由12 V电源供电。计算(a)总电阻，(b)总电流，(c)10 Ω电阻两端的电压。
For the parallel pair: 1/Rp = 1/10 + 1/15 = (3+2)/30 = 5/30 → Rp = 6 Ω.
并联部分的等效电阻：1/Rp = 1/10 + 1/15 = (3+2)/30 = 5/30 → Rp = 6 Ω。
Total resistance: Rtotal = Rp + 6 Ω = 6 Ω + 6 Ω = 12 Ω.
总电阻：Rtotal = Rp + 6 Ω = 6 Ω + 6 Ω = 12 Ω。
Total current from the battery: I = V / Rtotal = 12 V / 12 Ω = 1 A.
电池输出总电流：I = V / Rtotal = 12 V / 12 Ω = 1 A。
Voltage across the parallel block: Vp = I × Rp = 1 A × 6 Ω = 6 V. Since components in parallel share the same voltage, the 10 Ω resistor also has 6 V across it.
并联模块两端的电压：Vp = I × Rp = 1 A × 6 Ω = 6 V。由于并联元件两端电压相等，10 Ω电阻上的电压同样为 6 V。
3. Young’s Modulus and Material Selection | 杨氏模量与材料选择
Sample question: Explain why steel, which has a larger Young modulus than aluminium, is preferred for bridge girders.
样题： 说明为什么杨氏模量比铝更大的钢更适合用作桥梁大梁。
A larger Young modulus means greater stiffness – the material deforms less under the same load. Steel typically exhibits E ≈ 200 GPa, while aluminium alloys are around 70 GPa.
杨氏模量越大意味着刚度越高——在相同载荷下材料变形更小。钢的E值通常约为200 GPa，而铝合金大约为70 GPa。
When a girder bends, deflection is inversely proportional to E. A steel beam will therefore experience significantly less deflection for a given span and load, making it safer and more durable for long-span bridges.
当大梁弯曲时，挠度与E成反比。因此，对于给定的跨度和载荷，钢梁的挠度明显更小，使其在长跨桥梁中更安全、更耐久。
Although aluminium is lighter, its lower stiffness would require deeper or heavier sections to control deflection, often negating weight savings.
虽然铝材更轻，但其较低的刚度需要更深的截面或增加重量来控制变形，这往往会抵消自重优势。
4. Manufacturing Processes: Casting vs. Machining | 制造工艺：铸造与机加工
Sample question: State one advantage and one disadvantage of sand casting an engine block compared to CNC machining it from a solid billet.
样题： 与用CNC加工实心坯料相比，砂型铸造发动机缸体的一个优点和一个缺点是什么？
Advantage: Sand casting can produce complex internal cavities and near-net shapes in a single process, greatly reducing material waste and machining time. It is highly cost-effective for medium to large production runs.
优点： 砂型铸造能在单次工艺中制造复杂内腔和近净成形件，大幅减少材料浪费和加工时间。对于中大批量生产，成本效益极高。
Disadvantage: Cast surfaces are rougher and have lower dimensional accuracy, typically requiring additional machining on critical surfaces. Porosity and inclusions may also reduce mechanical properties compared to a machined forging.
缺点： 铸造表面较粗糙且尺寸精度较低，通常需要对关键表面进行额外加工。与机加工锻件相比，气孔和夹杂物还可能降低力学性能。
5. Mechanical Efficiency of a Lever System | 杠杆系统的机械效率
Sample question: A first-class lever lifts a load of 600 N. The effort applied is 150 N, and when the load moves up by 0.2 m, the effort moves down by 1.0 m. Find (a) mechanical advantage, (b) velocity ratio, and (c) efficiency.
样题： 一省力杠杆将600 N的负载抬起，施加的动力为150 N。当负载上升0.2 m时，动力端下降1.0 m。求(a)机械利益，(b)速比，(c)效率。
Mechanical advantage (MA) = Load / Effort = 600 N / 150 N = 4.
机械利益 (MA) = 负载 / 动力 = 600 N / 150 N = 4。
Velocity ratio (VR) = distance moved by effort / distance moved by load = 1.0 m / 0.2 m = 5.
速比 (VR) = 动力端移动距离 / 负载端移动距离 = 1.0 m / 0.2 m = 5。
Efficiency = (MA / VR) × 100% = (4 / 5) × 100% = 80%. The remaining 20% is lost to friction at the pivot.
效率 = (MA / VR) × 100% = (4 / 5) × 100% = 80%。其余20%消耗在支点处的摩擦上。
6. Potential Divider Circuits | 分压器电路
Sample question: A potential divider consists of a fixed resistor R1 = 2 kΩ in series with an LDR. The circuit is powered by a 9 V battery, and in bright light the LDR resistance falls to 1 kΩ. Calculate the output voltage across the LDR.
样题： 一分压器由固定电阻R1 = 2 kΩ与一个光敏电阻(LDR)串联构成，电路由9 V电池供电。在强光下，LDR的电阻降为1 kΩ。计算LDR两端的输出电压。
Use the potential divider formula:
Vout = Vin × (RLDR / (R1 + RLDR))
应用分压器公式：
Vout = Vin × (RLDR / (R1 + RLDR))
Substituting the values: Vout = 9 V × (1 kΩ / (2 kΩ + 1 kΩ)) = 9 V × (1/3) = 3 V.
代入数值：Vout = 9 V × (1 kΩ / (2 kΩ + 1 kΩ)) = 9 V × (1/3) = 3 V。
As light intensity increases, RLDR drops, causing Vout to decrease. This circuit could be used to trigger a night-light when the voltage rises above a threshold.
随着光照增强，RLDR下降，使Vout减小。该电路可用于当电压升高超过阈值时触发夜灯。
7. Gear Trains and Torque | 齿轮传动与扭矩
Sample question: A driver gear with 25 teeth meshes with a driven gear of 100 teeth. The input torque is 8 N m at 200 rpm. Assuming 100% efficiency, calculate the output speed and output torque.
样题： 一主动齿轮有25齿，与一100齿的从动齿轮啮合。输入扭矩为8 N·m，转速为200 rpm。假设效率为100%，计算输出转速和输出扭矩。
Gear ratio = driven teeth / driver teeth = 100 / 25 = 4 : 1 (speed reduction).
传动比 = 从动齿数 / 主动齿数 = 100 / 25 = 4 : 1（减速）。
Output speed = input speed / ratio = 200 rpm / 4 = 50 rpm.
输出转速 = 输入转速 / 速比 = 200 rpm / 4 = 50 rpm。
With 100% efficiency, power remains constant: Tin × ωin = Tout × ωout. Therefore, Tout = Tin × (ωin / ωout) = 8 N m × 4 = 32 N m.
在100%效率下功率守恒：Tin × ωin = Tout × ωout。因此，输出扭矩 Tout = 8 N·m × 4 = 32 N·m。
Note that torque increases by the same factor as speed decreases – a fundamental advantage of gearboxes.
注意扭矩以转速降低的相同倍数增大——这是齿轮箱的基本优势。
8. Tolerances and Clearance Fit | 公差与间隙配合
Sample question: A shaft is specified as 20.0 ± 0.05 mm. The mating hole is 20.1 ± 0.03 mm. Determine the maximum clearance, minimum clearance, and the type of fit.
样题： 一轴尺寸为20.0 ± 0.05 mm，配合孔为20.1 ± 0.03 mm。求最大间隙、最小间隙及配合类型。
Maximum clearance = largest hole – smallest shaft = (20.1 + 0.03) – (20.0 – 0.05) = 20.13 mm – 19.95 mm = 0.18 mm.
最大间隙 = 最大孔 – 最小轴 = (20.1 + 0.03) – (20.0 – 0.05) = 20.13 mm – 19.95 mm = 0.18 mm。
Minimum clearance = smallest hole – largest shaft = (20.1 – 0.03) – (20.0 + 0.05) = 20.07 mm – 20.05 mm = 0.02 mm.
最小间隙 = 最小孔 – 最大轴 = (20.1 – 0.03) – (20.0 + 0.05) = 20
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